DISCRETE APPLIED MATHEMATICS
EL.c~~IER Discrete Applied Mathematics 83 (1998) 135-155
Cycles in the cube-connected cycles graph *
Anne Germa a, Marie-Claude Heydemann b, Dominique Sotteau b,*
a ENST, 46, rue Burrault, Paris 13, Fratw
b LRI, (/A 410 CNRS, brit 490, C’rliccwitP de Paris-Sutf, Y14O_i Orsax, Frunw.
Received 27 July 1995; received in revised form 2 I February 1996; accepted 3 September I996
Abstract
In this paper we study the existence of cycles of all lengths in the cube-connected cycles graph and we establish that this graph is no far from being pancyclic in case n odd and bi-pancyclic in case n even. 0 1998 Elsevier Science B.V. All rights reserved.
1. Introduction
In this paper we study the existence of cycles of given lengths in the cube-connected
cycles graph CCC,, (see definition in Section 2). This graph was introduced by Preparata
and Vuillemin in 1981 ([6]) as a good alternate for the hypercube, having a fixed de-
gree equal to 3 and yet a small diameter compared to its number of vertices. This
graph was since then studied by a lot of people, who showed in particular, that it also
has good properties as far as communications are concerned. It was proved that CCC,,
contains a hamiltonian cycle, which is an interesting property to realise distributed al-
gorithms. However, its cycle structure is still not completely known, as to whether it
contains cycles of any given length. For n even, this graph is bipartite. For any n, n 3 2,
CCC, is a Cayley graph on the wreath product of Z/22 by ZfnZ [l]. Stong has given
a general construction for hamiltonian cycles in the case of Cayley graphs on wreath
product groups [8]. Rosenberg adapted the proof to CCC, [7]. He also proved the ex-
istence of some families of cycles in CCC, of length n2k - c(n - 2) (see Theorem 4. I
in Section 4). The problem of the existence of cycles of all possible lengths remains
open (i.e. cycles of all even length if n is even).
-’ The work was partially supported by GDR/PRC PRS. This work was partially done while the first author
was visiting LRI.
* Corresponding author. E-mail: [email protected].
0166-218X/98/$19.00 0 1998 Elsevier Science B.V. All rights reserved
PII SO 166-2 18X(97)00068-2
136 A. Germa et al. I Discrete Applied Mathematics 83 (1998) 135-155
In this paper we establish that the graph CCC, is not far from being pancyclic in
case II odd and even pancyclic (i.e. containing cycles of all even lengths) in case n
even. More precisely we prove the following result.
Theorem 1.1. The cube-connected cycles graph CCC3 contains cycles of length 3 and
all lengths between 8 and 3 x 23 = 24. CCC, contains cycles of length 4 and cycles
of all even lengths between 8 and 4 x 24 = 64. For n > 5, CCC,, contains cycles of
length n and of every even length between 8 and n2” except 10 and possibly n2” - 2.
Furthermore, for n odd, CCC,, contains cycles of every odd Iength between n •t 6 and
n2” - n - 2, and also cycles of length n2” - n $2.
The proof is divided into three parts corresponding to different sets of values of the
lengths 1 of the cycles, and different methods to prove the existence of such cycles.
First, in Section 3, using the hamiltonian cycle of the hypercube H(n) given by the
reflected Gray code, we directly construct cycles of lengths 1, for values of I roughly
between 3 x 2” and (n - 1) 2”. Then in Section 4, beginning with a construction of
cycles of CCC, given by Rosenberg [7] and deleting successively suitable vertices, we
construct cycles of lengths 1 for the highest values of 1. Finally, in Section 5, for n 3 3,
we construct cycles of the smallest lengths by induction.
2. Notation and definition
We consider simple graphs and use common terminology (for example, see [2]).
Let G = (V(G), E( G)) be a graph, with vertex set I’(G) and edge set E(G). We denote
by [u, v], u E V, v E V, the edge of G with ends u, v. We denote a path by the sequence
of its vertices, for example [uc, ut,. . . , uk]. The length of a path is as usual the number
of edges of the path (k in our example).
Binary strings are used as vertex labels or components of vertex labels for the classes
of graphs studied in this paper. For a binary string x = ~0x1 . - ..xi . . -x,_ I position 0
corresponds to the leftmost bit, position n - 1 to the rightmost bit. We use x(i) or
xoxt ...q. . . xn- 1 to denote the binary string obtained from x = xcoxi . . . xi . . .x+-l by
complementing the bit in position i. Similarly, x(i,j) and xc x1 . . xi. . . q. . .x,-l mean
x where bits i and j are complementary, and so on. We use ii to denote a string of
j i’s. We will abbreviate “the vertex with label x” to “vertex x”.
We denote by H(n) the n-dimensional binary hypercube. The graph H(n) has 2”
vertices which are labelled with the binary strings of length n. Two vertices of H(n)
are adjacent if and only if their labels differ in exactly one bit position.
Definition 2.1.
l The graph CCC, has n x 2n vertices, labelled (/,x), where 4 is an integer between
0 and n - 1, called the level of the vertex,’ and x is a binary string of length n,
’ In this paper, all arithmetic on indices and levels concerning CCC, is assumed to be module n.
A. Germa et al. I Discrete Applied Mathenmtics 83 11998) 13S155 137
called the YOM’ of x. Two vertices (/,x) and (e’, I)) are adjacent if and only if either
x = y and I/ - /‘I = 1, or C = P’ and y =x(e). In this last case, x and y differ in
exactly the bit in position C.
l Lruel C of CCC,, is the set of all vertices with level /. The edges joining two
vertices of the same level are called Ietiel-edges or H-edge’es.
l Ron, x of CCC,, is the set of all vertices with row x. The edges that connect (/,x)
to its neighbours (C + 1,x) and (G ~ 1,x) are called C-eck~es and form a cycle of
length n called the fundumentul cycle C(x) defined by x.
For II = 2, CCC,, is simply the cycle of length 8. So in this article we will only
consider the case n > 3.
3. Cycles of intermediate lengths
Let us recall that the cube-connected cycles graph of dimension n [6] is derived
from H(n) by replacing each vertex x of H(n) by a cycle of length n. Thus, a natural
idea to find cycles in CCC,, is to start from cycles of the hypercube. In this section,
we construct cycles of intermediate lengths using a hamiltonian cycle of H(n). Let us
first recall the construction of the hamiltonian cycle of H(n) induced by the symmetric
Gray code. We denote by V, the wrapped around sequence of vertices of this cycle.
It is constructed in a recursive way as follows. Starting in H( 1) with the sequence 0, 1,
the sequence %, of vertices of H(n) is obtained from the sequence %,_I of vertices
of H(n ~ 1) by writing the vertices of %‘,I_ I prefixed by 0 and then by the reverse of
% n_i, the vertices being now prefixed by 1. This can be written %n = O%,1_l, I%,,-1,
For example,
l for H(2) the sequence $91 is OO,Ol, 11,lO;
l for H(3) the sequence ??x is 000,001,011,010,110,111,101,100;
l for H(4) the sequence %?e is 0000, 0001, 0011, 0010, 0110, 0111, 0101, 0100, 1100,
1101, 1111, 1110, 1010, 1011, 1001, 1000.
Let x be any vertex of H(n). We denote by x+ and xP the vertices following and
preceding x, respectively, in the wrapped around sequence +Z, and we denote by U’(.Y)
the position of the bit on which x and x + differ. Let us recall that the positions of the
bits of a vertex x are numbered from 0 on the left to n - 1 on the right. We associate
to the vertex x of H(n) the two vertices of CCC, defined by ant(x)=(d(~~),x) and
succ(x) = (d(x),x). For example, in the case of H(4), if x = 0111, then xi = 010 1,
X- = 0110, d(x) = 2, ant(x) = (3,Olll) and .s~cc(~~) = (2,Olll). Thus for any vertex
x of H(n),
l the vertices succ(x)=(d(x),x) and ant(x+)= (d(x),x’) are adjacent in CCC’,,
through an H-edge in level d(x),
l the vertices ant(x) = (d(x-), ) x an succ(x)=(d(x),x) are on the same row, so that d
they can be joined, in CCC,,, by two different paths along the fundamental cycle
C(x) defined by x, the sum of the lengths of these two paths being equal to n.
138 A. Germa et al. I Discrete Applied Mathematics 83 (1998) 135-155
In order to get cycles of CCC,,, respecting the order of the vertices in the sequence
%‘n, we construct a new sequence by replacing each vertex x in +ZR by the sequence of
vertices of CCC, composed of ant(x), PI, succ(x), where P, is one of the two paths of
the fundamental cycle C(x) joining anr(x)=(d(x-),x) to succ(x)=(d(x),x), and z
the set of intermediate vertices of P,. For example, for n = 2, from %$ = {OO,Ol, 11, lo}
we get {(0,00),(1,00),(1,01),(0,01),(0, ll),(l, ll),(l, lO),(O, 10)). In this example
the paths P, are reduced to an edge and z is empty. In the general case, the length
of the new cycle is equal to 2”+’ plus the sum of the number of intermediate vertices
of the paths P,, i.e. Id(x) - n(x-)I - 1 or n - 1 - Id(x) - d(x-)I. All these numbers
are taken modulo II.
In the following lemmata we study the sequence of the values d(x), x E %?,,, and
the associated sequence Id(x+) - d(x)l, x E V,,, in order to calculate the lengths of the
cycles of CCC, (note that, since we will be interested in the sum of these values over
all x in %?,,, this is the same as to study the sequence of Id(x) - d(x-)().
Lemma 3.1. Let S,, denote the sequence of values d(x) for the vertices x of g,, and
S: be the sequence obtained from S,, by deleting the last element. Then
(i) For n=Z, S~=(l,O,l,O}, S~={l,O,l}.
(ii) For n>3, S,= {SA_, + l,O,Si_, + 1,0}, where SL_, + 1 denotes the sequence
obtained from SA_:_, by adding 1 to each element.
(iii) For n 32, each element of S, in odd position is equal to n - 1.
Proof. The proof is obvious by the inductive construction of %‘,, = OVn_ ~,l%?+i.
Having constructed the sequence &,-ii we obtain the sequence S, as follows: take
the n - 2 first terms of S-1 and add 1 to each of them (since we have introduced a 0
on the left of the strings, and then we have increased by one the “position of the bits”
used for d(x)), then take 0 (the corresponding vertices of H(n) differ in the first-left
position), and repeat twice the modified sequence. Cl
Lemma 3.2. Let D,, denote the sequence of the 2n dt#erences ld(x+) - d(x)l, x E $9,.
Then the following holds:
(i) For n=2, D~={l,l,l,l).
(ii) For n33, Dn={Di_,,n - 1,~ - l,Di_,,n - 1,n - l}, where Di_, is the sub-
sequence of D,,_l containing the 2”-’ - 2 first elements.
(iii) D, contains 2”-’ values equal to i for 1 d i bn - 2 and 4 values equal to
n - 1.
Proof. Since S, = {So_, + l,O,SL_, + l,O}, the second part of the sequence D, repeats
the first one. The 2”-’ - 2 first elements of D, are equal to the 2”-’ - 2 first elements
of D,_I. Since S,!_i begins and ends by n - 2, the 2+’ - 1 and 2”-‘th elements of
D, are n - 1. Part (iii) is easy by induction. 0
To prove Proposition 3.6 we need some technical lemmata.
A. Germu et al. IDiscrete Applied Mathemutics 83 11998) 135-155 139
Lemma 3.3. Let qi, 1 <id k, be given positive integers and consider the set oj’integers
E = {c,“=, ipi, O< p, 64;). Then E = (0, 1,. , c,“_, iqi} (i.e. E contains ull inteyers
,fLom 0 t0 I,“_, iqi).
Proof. The proof is by induction on k. It is clear for k = 1. Assume
k-l
Cipi, O<pi<qi I=1
] = {O,l,...,~iq~].
Let US consider some value of r such that 0 <r < C,“=, iq,. If r > kqk, we consider
r’ =r - kqx_, SO that r’< c,“r: iq,. By the induction hypothesis, r’ = cfrt ip, and
thus r = c,“zl ip, + kqk. If r < kqk the euclidean division of r by k gives r = kpk + rh
withrk=O orYk<k and pk<qk. Then r=s+kpk for somes,l<s<k-1 and pk<qk.
and this finishes the proof by taking pY = 1 and p, = 0 for i # k, s. 0
As a corollary of Lemma 3.3 we obtain the following result.
Lemma 3.4. Let qi, 1 <i<k, be given positive integers and let us consider the set of
integers E = {Ck=, 2ipi, 0 < pi <qi}. Then E = {0,2,. . ,2 CX=, iqi} (i.e. E contains
all even integers between 0 and 2 Ckzm, iq;).
In case of odd terms we get the following result.
Lemma 3.5. Let qi, 1 <i <k, be given positive integers with qo 32, and let us con-
sider the set of integers E = { cf=, (2i + 1 )pi, 0 < p; f qi}. Then E = (0, 1, , C,“_ ,
(2i+ l)q,} (i.e. E contuins nil integers between 0 und Ck=, (2i + 1)qi).
Proof. The proof is quite similar to the proof of Lemma 3.3, so we omit it. c!
The construction given above allows us to prove the existence of cycles of interme-
diate lengths in CCC(n) as specified in the next theorem.
Proposition 3.6. For n even, CCC,, contains cycles of’ all even lengths betcreen
lmin_ev = 3 x 2” - 2n/2’2 + 4 and I,,,-,, = (n - 1)2” + 2n.‘2+2 - 4. For n odd, CCC,,
contains cycles of all lengths between lmin__od = 3 x 2” - 3 x 2(“+‘)‘2 + 4 and lmax_od =
(n - 1)2” + 3 x 2(“+‘)j2 - 4.
Proof. As seen previously, the difference i = \d(x+) - d(x)1 for x E %n enables us
to construct two different paths 9, using C-edges, one containing i - 1 intermediate
vertices, the other containing n - 1 - i intermediate vertices.
By Part (iii) of Lemma 3.2 there exist 2”-’ paths P,, for 1 <i<n - 2, containing
either i - 1 or n - 1 - i intermediate vertices and 4 paths f’ containing either 0 or
140 A. Germa et al. I Discrete Applied Mathematics 83 (1998) 135-155
n - 2 intermediate vertices. Thus, the number of paths P, containing either i - 1 or
n - 1 - i intermediate vertices is equal to
0 2”-’ + 2’, for 2<i< [n/21,
l 2*-’ + 2’, for n odd and i = [n/21,
l 2nf2, for n even and i = n/2,
0 2”-’ $4 for i= 1.
For a given choice of paths P,, let us denote by pi, 1 <i< ln/2j, the number of
paths 9, containing i - 1 intermediate vertices in the constructed cycle of CCC,.
We split the proof into two cases, depending on the parity of n.
Let us recall that the length of a cycle of CCC, we have constructed is equal to
twice the number of vertices of the hamiltonian cycle of N(n) we started from, i.e.
2 x 2”, plus the sum of the number of intermediate vertices in the paths P, for all x
in ?Zfl.
Cuse 1: n odd. By considering all the possible choices of paths P, we see that in
CCC,, we can obtain cycles of lengths I with
1 = 2 X 2n + (2’-t + 4 - p1 )(n - 2)
1@1 + C [pi(i - 1) + (2,-j + 2’ - pi)(n - 1 - i)],
1=2
i 1421 I = 2 X 2” + (2+’ + 4)(n - 2) + C (2’-i + 2’)(n - 1 - i)
i=2 I
[
lfli21 - pl(n - 4 + C pdn - 29 ,
i=2 I for all O<pt<2”-’ +4 and 0<pj<2”-‘+2’, 2<i<dn/21.
From this we deduce that the minimum length Imin-od of this set of cycles is obtained
for pi = 2”-’ + 4 and pi = 2”-’ + 2’, 2 <i < [n/21, i.e., Zmin_od = 3 x 2” - 3 x T!(“+‘)/~
+ 4. We deduce also that the maximum length of this set of cycles is obtained for
pi = 0,l did Ln/2j, and is Imax-,,d =(n - 1)2” + 3 x 2(n+‘)/2 - 4.
To end the proof we only need to prove that, in fact, we obtain every intermediate
value between Irnm+,,j and Imax_od. Since n - 2i is odd, we can write the following
equality.
lni2J (n-3)/2
c pi(n - 2i) = c (2j + l)P(n-2j-l)/2.
i=l j=O
Using Lemma 3.5, we see that we obtain any intermediate value between Zmin-od and
hnax--od.
A. Genna et al. I Discrete Applied Mathematics 83 i 1998) I35 155 141
Case 2: n even. The proof is almost the same as in the case n is odd. CCC, contains
cycles of lengths
I = 2 X 2” + (2+-I + 4 - pl)(n - 2) + 2”“(n ~ 2)/2
+ C [pl(i - 1) + (2”-i + 2’ - pi)(/Z - 1 - i)],
i=2
r r1:2- I 7
1 = i 2 x 2” + (2’-’ + 4)(n - 2) + 2”/2(n - 2)/2 + C (2’+ + 2’)(n - 1 - i) r=2 I
n,‘Z-I
~ ! m(n
-
2) + C p,(n
- ,
i=2 24 1 for all 06 pl <2”-’ + 4 and 06 pi <2”-’ + 2’, 2 <i <n/2. From this we deduce
that the minimum length 1,,,_,, of this set of cycles is obtained for pI = 2”-’ + 4
and pI =2np’ + 2’, 2<i<nn/2 - 1, i.e., lmin-ev = 3 x 2” - 2”:2t2 + 4. We deduce
also that the maximum length of this set of cycles is obtained for p, = 0 and is
I,,,_,, = (n - 1)2” + 2n!2+2 - 4.
Since n - 2i is even we have
/1;2- 1 (n-4)/2
C p,(n - 2i) = C 2jP(t,p2j) 2
1=2 /=I
We use Lemma 3.4 to get every even value between l,,in_e, and I,,,_,, and, thus, to
finish the proof. 0
4. Existence of long cycles
Let us first introduce some notation specific to this part. In the notation of a vertex
(e,x), where x is a binary string of length n, we may need to specify only some of
the bits, in which case for example Oku will be the binary string of length n with the
first k bits equal to 0 and any binary string u of length n - k. Similarly, Ok1016 will
stand for Ok 101 0’-k-3. More generally, 6 will stand for a string of zeros of necessary
length to complete the length of the total string to n.
To prove the existence of long cycles, we first need to give a contruction due
to Rosenberg [7], since the proof of our result is based on it. Part of the proof of
the following proposition has been taken from Rosenberg’s report, with the necessary
changes. We will use this result later.
Proposition 4.1. For all n, the cube-connected cycles graph CCC,, contclins u qdr
qf length 1 Jbr the following value.7 qf’ I:
0 l=n,
l l=n2k-(n-2)cfork=20r3undO<c<2”,
142 A. Germa et al. I Discrete Applied Mathematics 83 (1998) 135-155
l l=n2k -(n - 2)c for 4<k<n and O<C<~~ - 2k-2-(kmod2).
In particular, CCC,, is hamiltonian.
Proof. The proof first concentrates on the case c = 0 in the expression for the length
1 for the contained cycle. Hence, we wish to establish the containment in CCC,, of
cycles of lengths r~2~ for all k E Z,+, - {l}.
The proof is an adaptation of the one from [8] that establishes that CCC,, has a
hamiltonian cycle, by showing that every graph in the following family of subgraphs
of CCC, has a hamiltonian cycle. For k E Zn+l, the graph CCC,” is the maximal
connected component containing vertex (0,6) of the induced subgraph of CCC,, on the
set of vertices
v, =&f z, x {xOn-k: x E z,“}.
Note that CCC,‘) IS obtained from CCC,, by deleting all level-edges of CCC, at levels
k, k + 1,. . . , n - 1 and then deleting all vertices and edges that are no longer accessible
from vertex (0,6); in particular, CCC,“’ IS (isomorphic to) the n-vertex cycle C,, and
CCC:“’ is identical to CCC,,. We let CCC,‘“’ inherit a level structure in the natural
way from CCC,.
We establish that every graph CCC,” has a hamiltonian cycle, by induction on k,
with two base cases. The base case CCC’;” will yield the desired result for all even
values of k; the base case CCCL” will yield the desired result for all odd values ofk.
Lemma 4.2. For all k E Z,,_, -{ 1 }, if CCC,, ‘k’ is hamiltonian, then so also is CCC~k’2’.
Proof. An illustration of the construction can be seen in Fig. 2, where a hamiltonian
cycle in CCC: is shown, constructed from four copies of a hamiltonian cycle in CCC:.
The figure is given with n = 5, but could easily be extended by straight lines to the
right in each row. Assume for induction that we are given a hamiltonian cycle %?
in CCCLk’. We extend the induction by traversing the hamiltonian cycle in CCCAk’
repeatedly. As an aid in describing the multiple traversals, we say that a traversal
proceeds up the cycle when it proceeds along the cycle in the increasing order of the
levels of its vertices, j,j + 1,j + 2,. and that the traversal proceeds down the cycle
when it proceeds along the cycle in the order j, j - 1, j - 2,. . of the levels of its
vertices, all addition being modulo n2k.
Implicit in the formula for pruning CCC,, to produce CCC’ik’ is the fact that one can
construct CCCjk+2’ by taking four copies of CCC:“! call them Copies 00, 10, 01, and
11, and interconnecting them so as to obtain a copy of CCCL’+‘). The interconnection
begins with a renaming of the row strings of the vertices of CCC,$k’ as indicated in
the following table.
In CCCik’ II In Copy 00 In Copy 10 In Copy 01 In Copy 11
x0-k II x0”-k IXloo”-k-2 ~X010”-k-2~X110”-k-*
A. Germa et al. IDiscrete Applied Mathematics 83 (19981 135-155 143
Now interconnect the four copies by adding to them the level-/z and level-(k + I)
edges of CCC, in just the way that makes the resulting graph isomorphic to
ccc:“+2’.
One can now trace out a hamiltonian cycle in CCC!k+2’ as follows. We refer freely
to the four copies of CCC;“’ that comprise CCCAki”‘.
1. Start at vertex (k, 0’) in Copy 00 of CCCAk’, and proceed up its hamiltonian cycle
until vertex (k + 1, 0’).
2. Cross from vertex (k + 1, 0’) in Copy 00 to vertex (k + 1, OkOIG) in Copy 01.
3. Starting at vertex (k + 1, OkOlo’) in Copy 01, proceed down its hamiltonian cycle
until vertex (k, OkOIG).
4. Cross from vertex (k, OkOIG) to vertex (k, Ok1 lo’) in Copy 11.
5. Starting at vertex (k, Ok1 lo’) in Copy 11, proceed up its hamiltonian cycle until
vertex (k + I, 0” 116).
6. Cross from vertex (k + 1, Ok 1 la) to vertex (k + 1, Ok 100’) in Copy 10.
7. Starting at vertex (k + 1, OklOG) in Copy 10, proceed down its hamiltonian cycle
until vertex (k, OklOG).
8. Cross from vertex (k, OklOG) to vertex (k, 0’) in Copy 00.
We claim that the above procedure does, indeed, specify a walk within CCCAkf2’,
i.e. that every prescribed step of the walk crosses just one edge of the graph. A facet of
this claim that is not completely evident resides in our procedure’s implicit exploitation
of the following property.
Property 1. For all k, every pair of vertices, (/,x) und (&+ 1 ,x), where L E Z,, - Zk and
x t ZT, appear consecutively in the hamiltonian cycle ji)r CCC:” thut our procedure
produces.
Property 1 assures us that the pairs of vertices, (/,x) and (G + 1,x), can be used
to interconnect cycles in copies of CCC:” m the way mandated by our procedure.
One verifies the property by noting that at the point when our procedure produces a
hamiltonian cycle for CCCAk’, the vertices of interest are of degree 2 in CCCik’, hence
must appear consecutively in the cycle.
The proof of Lemma 4.2 is now complete: the walk specified by our procedure
interconnects the hamiltonian cycles in the four copies of CCCAk’ in a way that yields
a hamiltonian cycle in CCCikt2’. 0
The proof of Proposition 4.1 when c = 0 is completed by establishing the base cases
of the induction.
Lemma 4.3. Both CCCA”’ and CCCj3’ are hamiltoniun.
Proof. Because CCC,$” IS (isomorphic to) the n-vertex cycle VRr it is hamiltonian.
Let us concentrate, therefore, on finding a hamiltonian cycle in CCC,$3’ (see
Fig. 1).
144 A. Germa et al. I Discrete Applied Mathematics 83 (1998) 135-1.55
Fig. 1. A hamiltonian cycle in CCC?)
We produce a hamiltonian cycle in CCCL3’ from copies of the hamiltonian cycle
in CCC;” in much the same way that we produced a hamiltonian cycle in CCC~k+2’
from copies of a hamiltonian cycle in CCCLk’ m Lemma 4.2, except that we need
eight copies of the “seed” cycle here, as opposed to the four copies that sufficed
there.
Let us take eight copies of the hamiltonian cycle in CCC~“‘, call them Copies
000,001,. . . , 111. Note that the vertices of the cycle comprise the set {(e, 0’) : & E Zn}.
Relabel the row strings in all copies of the cycle so that the vertices of Copy copy of
the cycle ((x,/I, y E Z2) comprise the set ((8, a/?$) : L E Zn}. Under this vertex labelling,
the vertex set of CCC’A3’ is just the union of the vertex sets of the eight cycles, and the
edges of CCCL3’ are the union of the edges of the cycles, plus the H-edges in levels
0, 1,2 of CCC,. Using the same notion of traversing a cycle by proceeding up the
cycle or down the cycle as we used in Lemma 4.2, we can now specify a hamiltonian
cycle in CCCL3’ as follows:
1. Start at vertex (1, 0000’) and proceed down the cycle, until vertex (2, 0000’).
2. Cross from vertex (2, 0006) to vertex (2, OOlG) in Copy 001.
3. Proceed from vertex (2, 0010’) in Copy 001 up the cycle, until vertex (1, 0016).
4. Cross from vertex (1, 0010’) to vertex (1, 0110’) in Copy 011.
5. Proceed from vertex (1, 01 lo’) in Copy 011 up the cycle, until vertex (0, 01 lo’).
6. Cross from vertex (0, 01 lo’) to vertex (0, 11 la) in Copy 111.
7. Proceed from vertex (0, 1 1 16) in Copy 111 down the cycle, until vertex (1, 11 la).
8. Cross from vertex (1, 1110’) to vertex (1, 1013) in Copy 101.
9. Proceed from vertex (1,lOlo’) in Copy 101 down the cycle, until vertex (2, 1016).
10. Cross from vertex (2, 1016) to vertex (2, 1000’) in Copy 100.
11. Proceed from vertex (2, 1003) in Copy 100 up the cycle, until vertex (1, 1006).
12. Cross from vertex (1, 1003) to vertex (1, 1106) in Copy 110.
13. Proceed from vertex ( 1, 1100’) in Copy 110 up the cycle, until vertex (0, 1100’).
14. Cross from vertex (0, 1100’) to vertex (0, 0100’) in Copy 010.
15. Proceed from vertex (0,0103) in Copy 010 down the cycle, until vertex (1,OlOa).
16. Cross from vertex (1, 0100’) to vertex ( 1, 0000’) in Copy 000.
The described walk constitutes a hamiltonian cycle in CCC!3’. The existence of this
cycle establishes the base case of the induction. 0
To extend our result to the entire claimed set of contained cycles, by allowing the
constant c to assume the given nonzero values, we need the following property.
Property 2. For ull values of x E Z;l, except 03u ,fiv k >3 odd, u E Z;‘-“, und O’u
ftir k >2 wen, u E Z;‘-2, whenever the Iralks thut define the cycles of 1enyth.s t12’
encomter u fundumental cycle C(x) qj’ CCC,, dejrzed by u gicen row string s, the_~
traoersr ull (M - 1 ) edges of C(x) other than the one that connects tlz~o of’ its udjuccnt
wrtices (/. x) and (P + 1, x) for some /.
If, for the appropriate values of x, we alter the walk so that in C(x) it traverses only
the edge connecting (d, x) and (P + 1, x) instead of the other n - 1 edges of C(x),
then the length of the entire traversed cycle is decreased by precisely n - 2. Pruning
the walk in this way for c row strings yields the generalized result where c’ can vary
between 0 and 2” - 2”-2 if k is even and 2” - 2”-’ if k is odd.
Property 2 is easy to verify by first noticing that it is valid for all possible rows
x if k = 2 or 3 (see the given construction in CCC!” and CCC,(,3’). Then in the given
procedure above, for k 22, the walk in CCC,, (‘Q) interconnects the hamiltonian cy-
cles in the four copies of CCC:” by breaking one edge in each of the four rows
O”OOG, 0” 106, Ok016 and Ok 1 lo’. Thus, the structure of the walks in the rows different
from these is not changed, and in particular, the structure of the hamiltonian cycle
of ccc:“’ m rows other than 02u is never altered when k is even and the structure
of the hamiltonian cycle of CCC, (” in rows other than 0’11 is never altered when k
is odd. 0
We will now show that we can decrease the length of each of the cycle obtained
in the previous proposition by a multiple of 6, and then by 4, or 8. An illustration of
the following lemma is given in Fig. 2.
Lemma 4.4. For any n, n 3 3, unJj k odd, 3 6 k <n, und uny r. I <r <(k ~ 1)/2 - 1.
the hamiltoniun cycle of CCCj;“’ d+ned aboce contains
(a) a puth of’ length 3 with extremities in level k : (k.G),(k - l,o’).(k - l,O”-’ lo’).
(k,O”-’ lo’),
(b) j;,r k>5, 2k-2r-2 paths of length 7 with extremities (2r + 1, 02’. lOuO”- ) und
(2v + 1,02’1 ltro”-“) , where u E ZiP2’.-‘. Each puth uses fi)ur C-edges hetnwn lwel
2~ and 2r + 1 on the rows 02’00u0”-k, 02” lOt10”~‘, 02’01uO”-h und 02' 1 IuO”-~. tuw
H-edges in level 2r and one H-edge in level 2r + 1.
Proof. Part (a) of the lemma can be verified directly for k = 3. CCCA3’ contains the
following path of length 3: (3,~),(2,~)),(2,001~),(3,001~). Also, in the
146 A. Germa et al. /Discrete Applied Mathematics 83 (1998) 135-155
00000
10000
OlOOQ
11000
00100
10100
01100
11100
cm10
10010
OlQlO
11010
00110
10110
01110
11110
00001
10001
01001
11001
Oil101
10101
01101
11101
0001 1
10011
01011
11011
co111
10111
01111
11111
0 1 2 3 4 0
t-
+
I _- _-
t- e-
t--
__
-
__
c___
Fig. 2. Existence of paths of length 3 and 7 in CCC,.
A. Germa et al. I Discrete Applied Mathematics 83 11998) 135-155 147
procedure for k >/5 to construct the hamiltonian cycle of CCCLk’ from four copies
of a hamiltonian cycle of CCCAk-*’ we join the vertices (k - l,o’) of copy 00 and
(k - l,O”-*OlG) of copy 01, which produces the expected path of length 3, [(k,G),
(k - l,o’), (k - 1,0k~2010’),(k,0k-2010’)] in CCCAk’.
The proof of part (b) uses induction on k. Let k = 2m + 1, m 22.
Let us recall that, for ka5, the construction of the hamiltonian cycle of CCC::’
uses four copies of a hamiltonian cycle of CCCAkM2’. The procedure induces two
paths of length 7 with extremities in level k - 2. Indeed, we join the extremities
(k - 2,o’) and (k - 2, Ok-* 100’) from the paths of length 3 specified in (a) (respectively,
in copies 00 and 10 of CCC (k-2)) This creates a path of length 7 with extremities .
(k-2,0ke3100’) and (k-2,0k-3110’). Similarly, we join the extremities (k-2,0k-*OlG)
and (k - 2, Ok-* 1 la) from the paths of length 3 specified in (a) (respectively, in copies
01 and 11 of CCCAkP2’ ). This creates a second path of length 7 with extremities
(k - 2,0kP3 1016) and (k - 2,0ke3 1116). In particular, starting from CCCL3’, we obtain
a hamiltonian cycle in CCC:” which contains two paths of length 7 with extremities
in level 3. Thus, part (b) of the lemma is satisfied for k = 5. Assume now that k 3 7,
and that the result is true for CCCLkf’, k’<k - 2.
Notice that the construction of the hamiltonian cycle of CCCAk’ does not alter any
of the edges of the hamiltonian cycles of the four copies of CCCLk-*’ except for edges
joining vertices between levels k - 2 and k - 1. Thus, the number of paths of length 7
of the type given in the lemma with extremities in levels 2r + 1 for r between 1 and
m - 2 is simply multiplied by 4 in the construction. And two new paths of length 7
appear between vertices of level k - 2. q
The following proposition is an immediate consequence of Lemma 4.4.
Proposition 4.5. For any n > 5 and any k odd, 5 <k <n, CCCLk’ and thus CCC,,
contain cycles of all lengths 1 with I= n2k - 62 j?w 0<~<(2~-? - 2)/3.
Proof. It is clear from the proof that the paths of length 7 from Lemma 4.4 are all
vertex disjoint. Also, the extremities of any of these paths are adjacent in CCC,!,k’.
Thus, replacing any of these paths by the edge joining its extremities, we decrease
the number of edges in the hamiltonian cycle of CCCik’ by 6. This can be done
independently for all the paths. The total number of these paths of length 7 is equal
to
r=n,- I s=m-2
c 2k-2r-2 =2 c 4”=2k-?3-2,
r=l s=o
with m = (k - 1)/2. 0
The following lemma can be proved exactly as Lemma 4.4, and thus we omit the
proof.
148 A. Germa et al. /Discrete Applied Mathematics 83 (1998) 135-155
Lemma 4.6. For any n 3 4, any k even, 2 <k d n, and any r, 16 r < k/2 - 1, the
hamiltonian cycle of CCCLk’ defined above contains
(a) a path of length 3 with extremities in level k: (k, G), (k - 1, a), (k - 1, Ok-’ lo’),
(k, Ok-’ lo’),
(b) for k 24, 2k-2r-’ paths of length 7 with extremities (2r,02r-’ 10~0”-~) and
(2r,02r-’ ll~O”-~), where u E Zi-*‘-‘. Each path uses four C-edges between level
2r - 1 and 2r on the rows 02’-‘00uO”Pk, 02r-’ 10~0”-~, 02’~101u0”-k and
O*+’ 1 ~uO”-~, two H-edges in level 2r and one H-edge in level 2r + 1.
Counting the number of independent paths of length 7, each of which again can be
replaced by an edge, we get the following.
Proposition 4.7. For any n, n 34, any k even, 4 d k <n, CCCLk’ and thus CCC,, con-
tains cycles of all lengths 1, with I= n2k - 6p, for 0 </I <(2k-’ - 2)/3.
Remark 4.8. Note that to produce the cycles in Proposition 4.5 we only altered edges
from the original hamiltonian cycle in rows 00~. Thus, using Property 2, we can still
prune each of the new cycles by replacing a path of length n - 1 by an edge in
each of the 2k - 2k-2 rows of labels starting with 01, 10 or 11 as we did before.
Similarly, to produce the cycles in Proposition 4.7 we only altered edges from the
original hamiltonian cycle in rows Ou. Thus, using Property 2, we can still prune each
of the new cycles by replacing a path of length n - 1 by an edge in each of the 2k-’
rows of labels starting with 1.
Proposition 4.9. For any n, n 3 5, and any k odd, 5 d k <II, CCCLk’ and thus CCC,
contains cycles of all lengths 1, with 1= n2k - 4 - 68, for 0 <p < (2k-’ - 4)/3.
Proof. In CCCAk’ consider the two copies 0 and 1 of CCCik-“. In each copy we
take one of the cycles obtained in Corollary 4.7 and we delete the path of length 3
(which exists by Lemma 4.6) with extremities in level k - 1. In CCCLk’ we then join
the remaining paths from the two copies with the H-edges of level k - 1, [(k - 1, 0’) (k - l,Ok-‘lo’)] and [(k - l,Ok-*10’),(k - l,Ok-*110’)]. •1
Proposition 4.10. For any n 34, and any k even, 4 <k <n, CCC;“’ and thus CCC,,
contains cycles OJ all lengths 1, with I= r12~ - 4 - 6a, for 0 <N ~(2~~~ - 4)/3.
Proof. Again, consider two copies 0 and 1 of CCC$k-‘) in CCCAk’. In each copy, we
take either one of the cycles obtained in Corollary 4.5 for k > 4, or the hamiltonian cycle
of CCC3’ if k = 4, and we delete the path of length 3 (which exists by Lemma 4.4) n with extremities in level k - 1. In CCC,$k’ we join the remaining paths from
the two copies with the H-edges of level k - 1, [(k - 1, G), (k - l,Ok-’ lo’)] and
[(k - 1,0k-210’),(k - LO”-*ll@]. Cl
A. Germa et al. I Discrete Applied Mathematics 83 (19981 13Sl.55 14’)
Remark 4.11. Notice again that in the previous two propositions, for any k34, the
construction does not alter rows beginning with 1.
Proposition 4.12. For any n, II 24, CCC,, contains cycles qf’ ull lengths 1 \r.ith I=
n2”-S-6;‘fijr 06y<2”-3-2.
Proof. Let us first consider the case where n is odd. From Proposition 4.10, used with
k=n-1, CCCj;“P” contains cycles of length 1’ = n2”-’ -4-6x for 0 < cx < (2”-’ -4)/3.
From Proposition 4.7, used with k =n - 1, CCC?-” also contains cycles of length
I” = n2nP’ -6/1 for 0 < fi < (2”-*-2)/3. It is not difficult to verify, from the construction
of the cycles, that each such cycle contains a path of length 3 with extremities in level
n - 1: [(n - l,010”p2), (O,OlO”-*), (0, llO”-*), (n - 1, llO”-*)I.
Consider the two copies 0 and 1 of CCC:-” contained in CCC,,, and take in one
of them a cycle of length 1’ and in the other a cycle of length I”. Delete the path of
length 3 mentioned above in each of these cycles, and join the two remaining paths
in CCC, using the H-edges in level n - 1: [(n ~ l,010flp2),(n - 1,010”~31)] and
[(n - I,1 lo”-*), (n - 1,l lone3 l)]. We obtain a path of length 1’ + 1” - 4. Thus CCC,,
contains a cycle of length n2” - 8 - 67 for 067 <(2”-’ - 2)/3 + (2n-3 - 4)/3. If n
is even, the proof is exactly the same using Corollary 4.5 and Proposition 4.9 with
k=n-1. u
Remark 4.13. Notice that in each of the two previous propositions the contruction has
altered two and only two rows beginning with 1.
We conclude this part with the following result.
Proposition 4.14. For any n, n 24, CCC, contains cycles of ull even lengths 1 with
(n - 1 )2n < I< n2” except possibly n2n - 2. Moreover, ,for any n odd, n 25, CCC,,
contains cycles of’ ull odd lengths 1’ with (n - 1 )2n < I’ ,< n2” ~ (n - 2) except possihl!,
n2” - n.
Proof. The proof follows from what precedes, using the three remarks to prune the
cycles obtained in the previous propositions in order to decrease their lengths by a
multiple of n - 2 as we did earlier. Together this implies that CCC, contains cycles
of lengths
n2” - c(n - 2) - 6cr - ~4 - ~‘8,
where O<c62”-‘-2 (from the remarks), e=O or 1, s’=O or 1, s+s’<l and O<x<
(2n-2 - 4)/3. Note that these are not the best bounds that could be used, but they are
valid for n odd or even and are sufficient for what we wish to conclude. Indeed, again
by a very rough calculus, we get cycle lengths down to at least n2n - (n - 2)2+’
which gives what we want since n 34. 0
IS0 A. Germa et al. I Discrete Applied Mathematics 83 (1998) 135-155
5. Existence of small cycles in CCC,,
We first recall some well-known properties of CCC,, that we will use in some
technical lemmata to construct small cycles by induction.
Fact 1. CCC,,+I contains two disjoint induced subgraphs each one isomorphic to the
graph obtained from CCC,, by deleting all edges between levels n - 1 and 0.
This results from the one-to-one mappings from V(CCC,) to V(CCC,+I ):
V,XOXl . . .X,-l) + (e,xoxl . . .x,-,0)
and
(e,x()x, . . .X,-l) --+ (e,xoxl . . ‘X,-l 1).
Fact 2. For any edges [(8,x),(8,x(/))] and [(/,y),(e,y(e))], O<&n-1, contained in
level e, there exists an automorphism of the cube-connected cycles CCC,, which sends
vertices (/,x), (e,x(/)) on vertices (L’, y), (e, y(e)), respectively, and do not change the
level of any vertex.
This property comes from the edge-transitivity property of H(n). Thus, if CCC,, has
a cycle of length I containing an edge in level /, for some e, 0 < 8 <n - 1, then CCC,,
contains a cycle of the same length through any edge contained in level /.
Definition 5.1. We say that a cycle is of type 0 in CCC,, if it is a cycle of CCC,
which contains no C-edge between two levels n - 1 and 0, but which contains at
least two H-edges [(n - l,x),(n - l,x(n - l))] and [(n - 1, y),(n - 1, y(n - l))] in
the level n - 1. We denote by Ye(n) the set of the lengths of the cycles of type 0
in CCC,.
Definition 5.2. Similarly, we say that a cycle is of type 1 in CCC, if it is a cycle of
CCC, which contains exactly one C-edge between the levels n - 1 and 0, but which
contains at least two H-edges in the level n - 1, these edges being not adjacent to the
previous C-edge. We denote by s(n) the set of the lengths of the cycles of type 1
in CCC,.
We first begin with the case n = 3.
Lemma 5.3. CCC, contains cycles of type 0 and lengths 8,12,14,16,20 and cy-
cles of type 1 and lengths 9,11,13, 15,17,19,21. It also contains cycles of lengths
3,10,18,22,23,24 which are neither of type 0 nor of type 1.
Proof. This result is not difficult to check. 0
A. Germa et al. I Discrete Applied Mathematics 83 11998) 135 -155 151
Lemma 5.4. rf’ CCC, contains a cycle of type 0 and length 1, then CCC,,+, contuins
cycles of lengths I,1 + 4,l f 6 and of type 0. If; furthermore, CCC,, contains u c.~~,le
qf’ type 0 and length I’, then CCC,+, contuins u cycle of lemgth 1 + 1’ + 4.
Proof. Notice that, by Fact 1, a cycle of type 0 in CCC, is also a cycle of type 0
in CCC,,+ 1. Consider a cycle % in CCC,, of length 1 and of type 0 as a subgraph of
CCC,, I using the one-to-one mapping
(/,x0x, “.X,_,) + (d,X()Xl “.X,_,O).
Let [(n - 1,x),(77 - l,x(n - l))] be one H-edge of this cycle lying in the level
n - 1. We extend this cycle to a cycle of length I+ 6 in CCC,+1 by deleting the edge
[(n- l,x),(n - l.x(n- 1))] and adding the path [(n- l,x),(n,x),(n,x(n)),(n- 1,x(n)),
(n - l,x(n - l,n)),(n,x(n - l,n)),(n,x(n - l)),(n ~ l,x(n - l)]. Notice that this new
cycle has no C-edge between levels n and 0 and has two edges in level n.
In order to obtain a cycle of length I+ 4 from %, assume, without loss of generality,
that cycle (6 contains in level n - 1 the two H-edges [(n - l,x),(n - l,x(n - l))] and
[(n-l,y),(n-l,y(n-l))] andthatvertices(n-l,x),(n-l.,v),(n~l,y(n-l)),(n-l.
x(n- 1) are placed in this order on %?. Delete from ?Z the two H-edges [(n- 1,x), (n- I.
x(n - l))] and [(n - 1, y),(n - l,y(n - I))]. Doing so we obtain two paths: PI of
extremities (77 - 1, y),(n - 1,x), P2 of extremities (n - l,y(n - l)),(n - l,x(f7 - I)), the
sum of the lengths of these paths being equal to I- 2. By Fact 1, CCC’,+t contains a
path Pi isomorphic to P2, disjoint from PI, and of extremities (n- 1,x(n)), (n- I, y(n)).
A cycle of length 1 + 4 is obtained by considering Pl,(n,x),(n,x(n)), Pi,(n, y(n)),
(n, y) (see an example in Fig. 3).
In order to obtain a cycle of length 1 + 1’ + 4 in CCC,,+ 1, consider ‘6 and a cycle
‘6’ of length 1’ and type 0. The idea of the construction is similar to the previous one.
“-I 0 n-l II
c 0 x I
l Y
I 5.
G l x(n-1)
\ l ~0-1)
l l
l l
l . l l
l l l l
Fig. 3. Construction of a cycle of length 1 + 4 in CCC,,_,
x
Y
x(n)
y(n)
152 A. Germa et al. I Discrete Applied Mathematics 83 (1998) i35-155
C
1 x
Gil Y
x(n-1)
1’
y(n-1) C’
n-l
Fig. 4. Construction of a cycle of length I + 1’ + 4 in CCC,+l.
Let [(n - l,x),(n - l,x(n - l))] be an H-edge of V and [(n - l,y),(n - l,y(n - l))]
an H-edge of W. Denote by Ps, P4 the paths obtained from 97, V’, respectively, by
deleting edges [(n - l,x),(n - l,x(n - l))], [(n - l,y),(n - l,y(n - l))], respectively.
By Facts 1 and 2, CCC,,+1 contains a path Pi isomorphic to P4, disjoint from Ps, and
of extremities (n - l,x(n)),(n - l,x(n - 1,n)). A cycle of length If I’+4 in CCC,+,
is obtained by considering Ps, (n,x), (n,x(n)), Pi, (n,x(n - 1, n)), (n,x(n - 1)). 0
The proof is easier to understand from Fig. 4.
Proposition 5.5. For any n 3 3, CCC, contains cycles of type 0 of all even lengths
between 8 and 3 x 2” - 4, except 10 and 3 x 2” - 6.
Proof. It follows directly from Lemma 5.4 by induction on n. By Lemma 5.3, Yo(3)
contains {8,12,14,16,20}. Assume that, for a given n33, 90(n) contains cycles of all
even lengths between 8 and 3 x 2” - 4 except 10 and 3 x 2” - 6. Using Lemma 5.4, we
first get that 9To(n + 1) contains all cycle lengths To(n). Then because 90(n) contains
3 x 2” -4 and all even values from 8 up to 3 x 2” -8 except 10, then To(n+ 1) contains
cycle lengths 3x2”-4+4+1=3~2~+1 for 1=0,2,8,12,14,...~,3~2”-8 and
also I= 3 x 2” - 4. So it remains to show that &(n + 1) contains cycles of lengths
3 x 2” - 6, 3 x 2” - 2, 3 x 2” + 4, 3 x 2” + 6 and 3 x 2* + 10. But it is not difficult to
get these values using Lemma 5.4 and starting with different values from Ye(n), since
3x2”-6=3x2”-12f6,3x2”-2=3x2”-8+6,3x2”+4=3x2”-8+8+4,
3x2”+6=3x2n-10+8+4and3x2”+10=3x2”-8+14+4. 0
A. Germa et al. I Discrete Applied Mathematirs 83 (1998) 135-155 153
The following lemma is similar to Lemma 5.4, but for cycles of type 1.
Lemma 5.6. If’ CCC,, contains a cycle of type 1 and length I’, then CCC,,+, contains
u cycle of type 1 for each length 1’ + 1,l’ + 7. If, furthermore, CCC, contains u cycle
of type 0 and length 1, then CCC,,+, contains u cycle of type 1 and length I+ 1’ + 5.
Proof. Let %? be a cycle of length 1 and of type 1 in CCC,,. Suppose that % contains
the C-edge [(n - 1,x),(0,x)] between levels n - 1 and 0 in CCC,. It is sufficient
to use Fact 1 and to replace this edge of CCC,, by the path in CCC,,,, of length 2
[(n - l,x),(n,x),(O,x)] to obtain a path of length 1’ + 1.
Any C-edge [(n - 1,x), (0,x)], x E Z;, used in a cycle of type 1 in CCC, is extended
to a path of length 2, [(n - l,x),(n,x),(O,x)] in the corresponding cycle of CCC,,,,
Otherwise, in order to obtain cycle lengths 1’ + 7 and I+ 1’ + 5, the constructions are
the same as that of Lemma 5.4 which give a path of length 1 + 6 and 1 + 1’ + 4,
respectively. 0
Using Lemma 5.6 and Proposition 5.5, we get the following analogue of Proposi-
tion 5.5 concerning the existence of cycles of type 1.
Proposition 5.7. For any ~22, CCC2, contains cycles of type 1 and all even lengths
between 2p + 6 and 3 x 22P + 2p - 6.
For any p3 1, CCG,+I contains cycles of type 1 und of all odd lengths bet\vren
2p + 7 and 3 x 22P+’ + 2p - 5.
Proof. We proceed by induction on p. By Lemma 5.3, Y,(3) contains {9,11, 13, 15,
17, 19,21}. Let us assume that, for some p 32, $ (2p - 1) contains all odd cycle
lengths between 2p + 5 and 3 x 22P-’ + 2(p - 1) - 5.
Since, by Proposition 5.5, Yb(2p - 1) contains all even cycle lengths between 8 and
3 x 22P-’ - 4, except 10 and 3 x 2 *P - 6, then, using Lemma 5.6, 5 (2~) contains all
even cycle lengths between 2p + 6 and 3 x 2 ‘P-l -4+3x 22P-’ +2p- 1 -6+ 1 +4
which is equal to 3 x 2 ‘P + 2p - 6. Now, usi n Proposition 5.5 and Lemma 5.6, we g
obtain from the values in Yc(2p) and Yt(2p) that .Tr(2p + 1) contains all odd cycle
lengths between 2p + 7 and 3 x 22P+’ + 2p + 1 - 6. 0
Before summarizing the results of this section, let us recall the following result
proving that CCC, does not contain some small cycles.
Proposition 5.8 (Heydemann et al. [5]). Every non-fundamental cycle of CCC, has
length at least 8. Every non-fundamental cycle of CCC5 has length 8 or at least 11.
rf’n 26, every non-fundamental cycle of CCC, has length 8 or length at least 12.
Proposition 5.8 shows, in particular, that, if n>5, then CCC(n) does not contain a
cycle of length 10. Thus, the result of this section is the following proposition.
154 A. Germa et al. IDiscrete Applied Mathematics 83 (1998) 135-155
Proposition 5.9. CCC, contains cycles of length 3 and all lengths between 8 and 24.
For n even, n 34, CCC,, contains cycles of all even lengths between 8 and 3 x 2”+n-6,
except 10. For n odd, n>,5, CCC, contains cycles of all odd lengths between n + 6
and 3 x 2” -I- n - 6 and cycles of all even lengths between 8 and 3 x 2” - 4 except 10
and 3 x 2” - 6.
6. Conclusions
The proof of Theorem 1.1 is now evident. First CCC, contains cycles of all possible
lengths except 4,5,6,7. Let us recall that fundamental cycles are of length n. Then
for any n, n 24, other cycle lengths are obtained as follows. Even length values I are
covered by
l Proposition 5.9, for 8 d 1 d3 x 2” - 8 except I= 10,
l Proposition 3.6, for 3 x 2” - 12<1 <(n - 1)2” + 12,
l Proposition 4.14, for (n - 1)2’ < 1 < n2” except possibly 1 = n2” - 2.
For n = 4, these results give cycles of all even lengths between 8 and 64 = 4 x 24
excepted 10 (in this case n2” - 2 = n2” - (n - 2) = 62 is obtained). But it is easy to
find in CCC4 a cycle of length 10:
K0,04),(3,04),(2,04),(1,04),(1,0100),(0,0100),
(0,l lOO),( 1,1 loo), (1, lOOO), (0, lOOO)].
Thus, CCC4 contains cycles of all even lengths excepted 6.
For n odd, n >5, odd length values 1 are covered by
l Proposition 5.9, for n+661<3 x2”- 1,
l Proposition 3.6, for 3 x 2” - 196Z<(n - 1)2” + 19,
l Proposition4.14,forI=n2”-n+2and(n-1)2”+1~Z~n2”-n-2.
Thus, in this article we have proved that the cube-connected cycles graph CCC,, is
almost pancyclic. Let us now consider the missing values of cycle length. By Propo-
sition 5.8 and Theorem 1.1, for CCC5 it remains to study the existence of cycles of
lengths5x25-5=155,5x25-2=158,5x25-1=159.
The possible values of length cycle 1 which are not covered by our constructions
for n 2 6 are 1= n2” - 2 and, in case n odd, odd lengths I, 13 13, for
- l<n-2,l=n+2,l=n+4,
- l=n2”-n, _ n2”-n+4<1<n2”- 1.
Notice that, since CCC, is a subgraph of the butterfly graph [4], our results apply
also to this graph.
Bondy “meta-conjectured” in [3] that almost any non-trivial condition on a graph
which implies that the graph is hamiltonian also implies that the graph is pancyclic.
One can ask whether like CCC,, every Cayley graph on a wreath product is almost
pancyclic or even pancyclic.
A. Germa et al. I Discrete Applied Mathematics 83 (1998) 135-155 155
Acknowledgements
We are grateful to A. Rosenberg for allowing us to copy part of the file from [7]
he had sent us. We also thank the referees for their helpful remarks.
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